Week 12 Exercise Inference 10 springs have been tested and have had their spring

Question

Week 12 Exercise
Inference 10 springs have been tested and have had their spring constant recorded. This data has been provided in a .csv file on Blackboard.

1. Using MATLAB, import the data using csvread and determine the sample mean and variance (using in-built functions is fine).
2. Determine the 95% confidence interval for the spring constant of the springs.
3. The spring manufacturer claims that the springs have a spring constant of 10kN/m. Conduct a hypothesis test to see if there is sufficient evidence to suggest that this spring constant is incorrect. Use the comparison between Ttest and tN -1,1 - alpha/2 for your hypothesis test.
4. If the sample mean and variance stayed the same, but there were 50 springs, calculate the new 95% confidence intervals and reconduct a hypothesis test. Explain why the confidence interval changes.

Explanation / Answer

Solution

Let X = spring constant in (kN/m). Then, X ~ N(µ, 2).

Part (1)

Given sample data is as follows.

Using Excel Functions, sample size = n = 10;

sample mean = 9.81538 and sample variance = 0.093165 ANSWER

Part (2)

95% Confid3nce Interval for spring constant = {Xbar ± (s/n)(t/2)}, where Xbar = sample mean, s = sample standard deviation, n = sample size and t/2 = upper /2 percent point of t-distribution with degrees of freedom = n – 1. Given 95% Confidence Level, /2 = 2.5 and t9, 0.025 = 2.262.

So, 95% Confid3nce Interval for spring constant   

9.81538

±

0.066646

Lower Bound =

9.748734

Upper Bound =

9.882026

ANSWER

Part (3)

Claim: Springs have a spring constant of 10 kN/m

Hypotheses:

Null   H0:    µ   =   µ0   =

10

Alternative   HA: µ

10

Test Statistic:

t = (n)(Xbar - µ0)/s where

n = sample size =

10

Xbar = sample mean =

9.81538

µ0   (given)    =

10

s = sample standard deviation = 0.093165

    So, tcal =

-6.266513

   |tcal| =

6.266513

Distribution and Critical Value

Under H0, t ~ t with DF = n - 1   =

9

Taking =

0.05

Critical Value, tcrit = upper

/2 % point of tn-1 = 2.262

Decision Criterion (Rejection Region)

Reject H0 if |tcal| > tcrit

Since |tcal| > tcrit, H0 is rejected

Conclusion:

Given data suggest that the claim is not true.

DONE

  10.051,  

  9.5551,  

  9.806,  

  9.9991,  

  10.191,  

  10.259,  

  9.8472,  

  9.4386,  

  9.4493,  

  9.5575  

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